How do you solve $log_7 7(x+1) + log_7 (x-5)=1$ ?

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Answer 1 · 2015-11-01T10:03:09

$x = 2 ± 2sqrt 2$

$log a + log b = log (ab)$ , then
$log_7 [7(x+1)(x-5)] = 1$

$log_b a = c \Rightarrow b^c = a$ , then
$7^1 = 7(x+1)(x-5)$

$1 = (x+1)(x-5)$ , continue with Baskara

How do you solve log_7 7(x+1) + log_7 (x-5)=1log_7 7(x+1) + log_7 (x-5)=1?